Question

The foci of hyperbola $9x^2-16y^2+18x+32y=151$ are

Answer 1

$9x^2-16y^2+18x+32y=151$

$(9x^2+18x)-(16y^2-32y)=151$

$\Rightarrow ({\left(3x\right)}^2+2\times 3x\times 3+3^2-3^2)-({\left(4y\right)}^2-2\times 4y\times 4+4^2-4^2)=151$ by completing the square method

$\Rightarrow {(3x+3)}^2-9-{(4y-4)}^2+16=151$

$\Rightarrow {(3x+3)}^2-{(4y-4)}^2=151-7$

$\Rightarrow {(3x+3)}^2-{(4y-4)}^2=144$

$\Rightarrow 9{(x+1)}^2-16{(y-1)}^2=144$

$\Rightarrow \frac{9{(x+1)}^2}{144}-\frac{16{(y-1)}^2}{144}=1$ by dividing both sides by $144$

$\Rightarrow \frac{{(x+1)}^2}{16}-\frac{{(y-1)}^2}{9}=1$ is the equation of the horizontal hyperbola.

center$=(-1,1)$

We have $a=4$ and $b=3$

$c^2=a^2+b^2=4^2+3^2=16+9=25$

$\therefore c=\sqrt{25}=\pm 5$

Foci$=(-1\pm 5,1)$

$\therefore$Foci$=(-1+5,1)$ and $(-1-5,1)$

Hence Foci$=(4,1)$ and $(-6,1)$